Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $k = \dfrac{y^2 - 15y + 54}{10y + 20} \times \dfrac{y + 2}{9y^2 - 81y} $
Solution: First factor the quadratic. $k = \dfrac{(y - 9)(y - 6)}{10y + 20} \times \dfrac{y + 2}{9y^2 - 81y} $ Then factor out any other terms. $k = \dfrac{(y - 9)(y - 6)}{10(y + 2)} \times \dfrac{y + 2}{9y(y - 9)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (y - 9)(y - 6) \times (y + 2) } { 10(y + 2) \times 9y(y - 9) } $ $k = \dfrac{ (y - 9)(y - 6)(y + 2)}{ 90y(y + 2)(y - 9)} $ Notice that $(y + 2)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ \cancel{(y - 9)}(y - 6)(y + 2)}{ 90y(y + 2)\cancel{(y - 9)}} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $k = \dfrac{ \cancel{(y - 9)}(y - 6)\cancel{(y + 2)}}{ 90y\cancel{(y + 2)}\cancel{(y - 9)}} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $k = \dfrac{y - 6}{90y} ; \space y \neq 9 ; \space y \neq -2 $